Solve the equation. $\dfrac{dy}{dx}=y^{-1}x^{-3}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\pm\sqrt{\dfrac{x^4}{2}+C}$ (Choice B) B $y=\pm\sqrt{\dfrac{x^4}{2}}+C$ (Choice C) C $y=\pm \sqrt{-\dfrac{1}{x^2}}+C$ (Choice D) D $y=\pm \sqrt{-\dfrac{1}{x^2}+C}$
We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=y^{-1}x^{-3} \\\\ \dfrac{dy}{dx}&=\dfrac{x^{-3}}{y} \\\\ y\,dy&=x^{-3}\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} y\,dy&=x^{-3}\,dx \\\\ \int y\,dy&=\int x^{-3}\,dx \\\\ \dfrac{y^2}{2}&=-\dfrac{1}{2x^2}+C_1 \\\\ y^2&=-\dfrac{1}{x^2}+C \\\\ y&=\pm \sqrt{-\dfrac{1}{x^2}+C} \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\pm \sqrt{-\dfrac{1}{x^2}+C}$